What the value of p and q:

4p + 7 + 3qi = (p -qi) - 2(1 +5i)

okay. this is easy...but I cant help to bother with 'i' so yeah.

Can anyone show me the solution?

the answer is p = -3 and q = - 5/2

- Mar 3rd 2014, 06:23 AMpsitthis is simple complex numbers question...can show me the solution?
What the value of p and q:

4p + 7 + 3qi = (p -qi) - 2(1 +5i)

**okay. this is easy...but I cant help to bother with 'i' so yeah.**

Can anyone show me the solution?

the answer is p = -3 and q = - 5/2 - Mar 3rd 2014, 06:28 AMromsekRe: this is simple complex numbers question...can show me the solution?
If you were looking for a way to make sure we don't help you, you nailed it.

You can't be bothered with "i" ? Why bother taking the course then?

If this is a language problem I apologize but really... you are basically telling us you can't be bothered to do your homework and that we should do it for you.

Wrong. - Mar 3rd 2014, 06:29 AMHartlwRe: this is simple complex numbers question...can show me the solution?
Gather the real and imaginary parts together in the form a+bi = 0 and equate a and b to 0.

- Mar 3rd 2014, 08:22 AMpsitRe: this is simple complex numbers question...can show me the solution?
omg. what's going on here?(Doh)

language misunderstood? Yeah, English is not my first language.

bother like, as the question ask for p and q....yet I cant stop thinking of the 'i' in the question?

it paired with number and with q too....make me thinking what should I do with that letter 'i'.(Giggle)

should it be solved first, to find q and p? yeah,yeah I am low at Mathematics. btw, that's what I mean by I cant help to bother it. Even it's not asked to be solved. Sorry my language.

oh...okay...@Hartlw...thanks...

a + bi = 0 true......

I did some question it that form...

and when I met this question...

in this 4p + 7 + 3qi ... it became three part now...oh...I m lost.

I did try make the other equation, at the right, have '7' too...haha...

Goodness,....omg...I really hope you guys are cool and helpful too. Your readers are worldwide. Some of us are very very poor in Math and language too, like me.

oh btw...I almost 17 years off secondary school now...miss the homework day...(Crying)

just me try to catch up with Matriculation Mathematics here in Malaysia...and with secondary school too...

it was taught in Malay language in 1997 and less advanced compared to today here...

I search for appropriate board here..so pre-university might be right place...I plan to read and study all the comments here...and grow here too...but somehow...now ... I think...I m at wrong place?

or should I be in other board?

Am I wrong? - Mar 3rd 2014, 08:41 AMromsekRe: this is simple complex numbers question...can show me the solution?
ok, let's start again then.

$4p + 7 + 3q \imath=p-q \imath-2(1+5\imath)$

like Hartlw said just split things into the parts multiplied by $\imath$ and those not.

$4p+7=p-2$

$3p=-9$

$p=-3$

$\imath 3q=\imath(-q-10)$

$3q=-q-10$

$4q=-10$

$q=-\frac{10}{4}=-\frac{5}{2}$ - Mar 3rd 2014, 08:47 AMHartlwRe: this is simple complex numbers question...can show me the solution?
Don’t feel bad, mathematicians don’t “understand” i either. They define it as i

^{2}=-1, and then__define__a new set of “complex” numbers a+bi, or (a,b), with certain properties.

Most of what passes for “understanding” in mathematics is memorized definitions, which are usually omitted.

Google “origin of complex numbers” for some background. - Mar 3rd 2014, 12:17 PMHartlwRe: this is simple complex numbers question...can show me the solution?
Sorry psit. I answered your OP but I didn’t give an explanation.

Why does a+bi=0 imply a=0 and b=0? It’s not a matter of “understanding,” it’s simply a matter of having seen and remembered one of the defining properties of the complex number a+bi:

__Definition__: a+bi=0 if and only if a=0 and b=0.

If you go into mathematics, you'll run into a lot of this type of thing. Unless you're willing to accept things without "understanding," and just memorize them, it can be quite frustrating. - Mar 4th 2014, 04:24 AMpsitRe: this is simple complex numbers question...can show me the solution?
**Oh thanks romsek!**(Talking)

It's about arrange the equation..hmm...wow...relief

that need practice to me...I cant see it before...

thank you again! You made it simple to understand..~(Clapping) - Mar 4th 2014, 04:28 AMpsitRe: this is simple complex numbers question...can show me the solution?
ah..wow. That complex number origin is very helpful to know..oh thanks Hartlw...

uuuu...so there's a condition to that a+bi=0...hmmm...(Happy)

Hmmm...yeah...well...maybe as freshie,(Talking)

Im willing to accept without completely 'understanding'then...

Thank you again Hartlw(Wink) - Mar 4th 2014, 06:23 AMDevenoRe: this is simple complex numbers question...can show me the solution?
Hartlw makes a valid point: why should one accept that:

a+bi = 0 implies a = b = 0 (besides it being "true by definition")?

Note we are using the symbol "0" in 2 different ways here: in the equation a+bi = 0, we mean 0 is a COMPLEX number, in the equations a = b = 0, we mean 0 is a REAL number.

So what do we MEAN by the complex number 0?

We mean some complex number z' = a+ib such that, for any complex number z = x+iy, that:

z + z' = z' + z = z (in other words we want z' = 0 (the "complex 0") to be an additive identity for complex numbers).

So let's look at the equation:

z + z' = z in terms of real and imaginary parts:

the left side is:

(x+iy) + (a+ib)

and the right side is:

x+iy.

We are taking these two to be equal, so:

(x+iy) + (a+ib) = x+iy.

Now, how do we add complex numbers together? We add their real parts and imaginary parts separately. So the left side becomes:

(x+iy) + (a+ib) = (x+a) + i(y+b). So our equation is now:

(x+a) + i(y+b) = x + iy.

Since this is the same complex number on both sides, it has the same real and imaginary parts. Thus:

x+a = x

y+b = y

These are equations involving REAL numbers, and using arithmetic from the real numbers, we find that:

a = x-x = 0

b = y-y = 0, so that the complex number "0" is actually 0+i0, which makes a certain amount of sense. - Mar 4th 2014, 07:01 AMHartlwRe: this is simple complex numbers question...can show me the solution?
Everything in previous post is simply a consequence of the definition (Field) of a complex number, including addition. There are no "explanations."

The only significant contribution to "understanding" would be why the combination a+bi and the particular definitions (properties), field, were chosen in the first place. You would have to look into "The History of Mathematics," which I believe would show the origins in the attempt to solve polynomials. - Mar 4th 2014, 10:08 AMromsekRe: this is simple complex numbers question...can show me the solution?
- Mar 4th 2014, 12:51 PMHartlwRe: this is simple complex numbers question...can show me the solution?
You're kidding. Surely we can assume the definition of a complex number.

I note that in addition to a+bi=0 iff a=0 and b=0, you do need the definition of addition of complex numbers, which nobody pointed out. The rest is elementary arithmetic- that has to be explained? - Mar 4th 2014, 12:53 PMDevenoRe: this is simple complex numbers question...can show me the solution?
The complex numbers were slow in being accepted as "legitimate algebraic objects" (the term "imaginary" for multiples of i indicates why this might be so). In fact, it was not until the graphical representation afforded by the Argand plane was introduced that most objections were at least quelled.

It had long been known that an "addition" could be done in the Euclidean plane by adding "coordinates" (such an operation was initially called "translation" and is geometrically evidenced by the operation of "moving the origin to a new place"). Equality of 2 points was held to be only when their coordinates were equal (this seems more obvious now, but at one time, it had to be FORMULATED as a definition).

Regarding (1,0) as the real number "1" and (0,1) as the "imaginary" number i, thus settles any qualms we might have about ADDING (or subtracting) complex numbers, we have a "real part" (the x-coordinate) and an "imaginary part" (the y-coordinate).

We also, from this view, have a perfectly good idea of what it means to "stretch" a complex number by a real number: we simply set r(x + iy) = (rx) + i(ry) (or, in plane coordinates: r(x,y) = (rx,ry)).

However, this gives us no "good idea" as to what MULTIPLICATION betwixt two complex numbers might mean. And one cannot have a field without some form of multiplication.

Now, one could simply DEFINE what one wants (a + ib)(c + id) to be, but this is not very illuminating: WHY do we want to define it that way?

One answer might be: we want distributivity to hold, so, we wish:

(a + ib)(c + id) = a(c + id) + (ib)(c + id) = ac + a(id) + (ib)c + (ib)(id).

To simplify this further, we would need to decide whether or not the imaginary number i "commuted" with real numbers. If we decided "yes", we then have:

(a + ib)(c + id) = ac + (ad)i + (bc)i + (bd)(i^{2}).

Now our original numbers were "polynomials in i" of degree 1, but now we have a "polynomial in i" of degree 2, which violates closure (we want our products to wind up in the same set we started out in). Of course, we could just DECLARE i^{2}= -1, but this seems a bit like "cheating", we're "rigging the game" to get what we want. So let's pretend we don't know what "multiplying by i" does just yet.

One observation that can be made, is that the mapping "multiplication by a real number r" in terms of "points in the plane" is just multiplication by the 2x2 matrix rI =

$$\begin{bmatrix}r&0\\0&r\end{bmatrix}$$

(such a matrix is called a "dilation matrix").

So we might think that "multiplying by i" might involve a matrix that takes "1" (that is, the point (1,0)) to "i" (that is, the point (0,1)).

If we write this as a matrix equation, this is:

$$\begin{bmatrix}a&b\\c&d \end{bmatrix} \begin{bmatrix}1\\0 \end{bmatrix} = \begin{bmatrix}0\\1 \end{bmatrix}$$

This gives us: a = 0, c = 1.

One matrix that does this is "rotation by 90 degrees counter-clockwise" or:

$$J = \begin{bmatrix}0&-1\\1&0\end{bmatrix}$$.

Note that J^{2}= -I (where I is the 2x2 identity matrix). Since we have identified rI with the real number r, J^{2}is identified with the real number -1.

This suggests that we think of complex numbers as "rotation-dilations of the plane", represented as the matrices of the form:

$$\begin{bmatrix}a&-b\\b&a \end{bmatrix}$$

"a" is the "stretchy" part (dilation), and "b" is the "turn-y" part (rotation).

Let's then, compute, the result of applying aI + bJ to a complex number c+id:

$$\begin{bmatrix}a&-b\\b&a \end{bmatrix}\begin{bmatrix}c\\d \end{bmatrix} = \begin{bmatrix}ac-bd\\ad+bc \end{bmatrix}$$

This suggests that we DEFINE:

(a + ib)(c + id) = (ac - bd) + i(ad + bc)

and what we know about 2x2 matrices allows us to conclude that we have, at least, a ring.

Furthermore, if a = c = 0, and b = d = 1, the above simplifies to:

(i)(i) = (0+i)(0+i) = (0-1) + i(1-1) = -1 + i0 = -1, as we hoped for, earlier.

Now (c + id)(a + ib) = ca - db + i(da + cb), but since a,b,c, and d are all real numbers, we have ac = ca, db = bd, da = ad, cb = bc, so that:

(c + id)(a + ib) = (a + ib)(c + id), so our ring is commutative (this turns out not to be the case when we try to do this trick AGAIN by replacing real numbers with pairs of complex numbers, which gives the quaternions, which are "strange").

Now, all we lack is to find an inverse for any non-zero complex number a+ib, to ensure we have a field.

There's a "trick" to this:

(a + ib)(a - ib) = (a + ib)(a + i(-b)) = a^{2}+ b^{2}+ i(a(-b) + ab) = a^{2}+ b^{2}+ i0 = a^{2}+ b^{2}.

Now as we saw earlier a+ib = 0 (as a complex number) means a = b = 0 (as real numbers), so if a+ib is NOT zero (as a complex number) then at least ONE of a,b is NOT zero.

This means that the real number a^{2}+ b^{2}is non-zero, so we can safely DIVIDE by it, to form the (real fraction) 1/(a^{2}+ b^{2}).

And THIS means that:

(a+ib)(([1/(a^{2}+ b^{2})](a - ib)) = (a^{2}+ b^{2})/(a^{2}+ b^{2}) = 1, so:

(1/a^{2}+ b^{2})(a - ib) is an inverse for a + ib.

*****************

It is a fair question to ask: why would we do this?

The simplest answer I can think of comes from the quadratic formula, where one has for:

ax^{2}+ bx + c = 0

the solution:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

If b^{2}- 4ac < 0, one is often told "there is no solution" which is only HALF-true. There is no REAL solution, but if we allow complex numbers there is always a complex solution.

And, since the complex numbers form a field, we can "do algebra" with them in largely the same way as with "ordinary" (real, or rational) numbers.

It turns out, that in the physical world, some quantities act as if they are obeying "complex number rules" but we can only see the "real shadows" of this behavior (one example of this is the behavior of capacitors when subjected to alternating current). - Mar 4th 2014, 12:54 PMromsekRe: this is simple complex numbers question...can show me the solution?